The power plant consists of boiler and turbogenerator halls, electrostatic preci

Question

The power plant consists of boiler and turbogenerator halls, electrostatic precipitators, maintenance buildings, offices, roads, and parking, covering an area of about 75 hectares (750,000 m2). (A hectare is equal to a square 100 m on a side.) Adding the annual area stripped at the mine, the total area used each year by the entire facility is (199,000 + 750,000 =) about 950,000 m2 and the resulting overall power density is for the plant is 960 MW/950,000 m2 = 960 x 106W/950,000 m2 = 1010 W/m2

Part 2. Coal-fired power plant #2 is located in Georgia. It is not as efficient as plant #1. It, too, has a rated maximum power output of 1.2 GW, but it operates at a lower capacity factor of 70% and has a conversion efficiency of only 33%. It burns lower quality (less energy dense) coal with a higher content of sulfur and ash, that must be transported a long distance from the mine to the plant. It is not situated near a river and so must recycle cooling water in large towers. Sulfur and ash must be removed from the stack gases, and the ash must be disposed of. 2) What is the average power output of Plant #2?

3) What is the annual energy output of Plant #2, in PJ?

4) How much primary energy (in the form of coal) does Plant #2 require to operate each year, in PJ?

Explanation / Answer

2) The maximum power output can be 1.2 GW. But we have to add 0.70 capacity factor and 0.33 efficiency factor.

So after capacity factor of 0.7 we get, 1.2*0.7 =0.84 GW and after burning 0.84 GW energy coal we will get. 0.84* 0.33 = 0.2772 GW of energy.

3) 0.2772 GW of energy is 277200000 watts and 1 watt equals to 1J/second

so there will be 2772 PJ/second

4) The coal has around 30 MJ/kg energy, we need 2772 PJ so we will need,

277200000/30 = 9,240,000 kg of coal annually

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